Question

prove that root 3 is an irrational number​

Answer 1

Step-by-step explanation:

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

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Answer 2

Answer:

let us assume that root 3 is rational number .

Step-by-step explanation:

so , √3 = a / b ( HCF of a and b is equal to 1 ) .

squaring on both sides ,

(√3) square = a square / b square

we get , 3 = a square / b square

3b square = a square .... equation 1

b square = a square / 3

a square divides 3

a divides 3

let a = 3c ( where c is any integer )

substituting in equation 1

3b square = a square

now , 3b square = (3c ) square

3b square = 9c square

( dividing both sides by 3 ), we get

b square = 3c square

b square / 3 = c square

b square divides 3

b divides 3

therefore , a and b have 3 as their common factor other than 1 . this is contradiction to our assumption.

so , √ 3 is irrational .

Hope this helps you ..